Question

Find parametric equations for the tangent line to the helix with parametric equations x = 5 cos(t) y = 4 sin(t) z = t at the point (0, 4, π/2). SOLUTION The vector equation of the helix is r(t) = 5 cos(t), 4 sin(t), t , so r'(t) = . The parameter value corresponding to the point (0, 4, π/2) is t = , so the tangent vector there is r'(π/2) = . The tangent line is the line through (0, 4, π/2) parallel to the vector , 0, , so by these equations its parametric equations are the following. x = y = 4 z =

Answer 1

The parametric equation of the tangent line is:

x= -5t

y= 4

z= t +
`(\pi)/(2)`

Explanation:

The vector r(t)=[5 cos(t) , 4 sin(t) , t]

Hence, the equation of the tangent line is the derivative of the vector r(t):

r'(t)= [-5 sin(t) , 4 cos(t) , 1]

The tangent line is given by:

x= 0 + t·[-5 sin(t)]

x= 4 + t·[4 cos(t)]

y=
`(\pi)/(2)`+t

Given the point (0, 4, π/2), t=
`(\pi)/(2)`

So we replace this value in the equation of the derivative equation:

`r'(t)=[-5 sin((\pi )/(2)) , 4 cos((\pi )/(2) ) , 1]=[-5,0,1]`

The parametric equation of the tangent line is:

x= -5t

y= 4

z= t +
`(\pi)/(2)`